# $a^2 <10^{\sqrt{a}}$ for $a\geq 2$.

How to show $$a^2 <10^{\sqrt{a}}$$ for $$a\geq 2$$ and $$a \in \mathbb{N}$$?

Should I try considering a new function which is the difference and then differentiating it?

I could solve it using two case when $$a \in [10^{2m}, 10^{2m+1})$$ or $$[10^{2m-1}, 10^{2m})$$. Is there any other elementary way?

Any help would be appreciated. Thanks in advance.

## 2 Answers

That inequality is equivalent to $$\ln a < \sqrt{a} \frac{\ln 10}{2}$$ because of the fact that $$t \to \ln t$$ is strictly increasing.

Define $$f(a)=\ln a$$ and $$g(a)=\sqrt{a} \frac{\ln 10}{2}$$ for $$a \geq 2$$.

We want to show that $$f(a) for all $$a \geq 2$$.

Noting that $$f'(a)=\frac 1 a \leq \frac {1}{\sqrt{a}}\frac {\ln 10} {4}=g'(a)$$ holds if and only if $$\sqrt{a} \geq \frac{4}{\ln 10}$$, define $$\phi(a)=g(a)-f(a)$$.

Due to the fact that $$\phi'(a) \geq 0$$ if and only if $$a \geq (\frac{4}{\ln 10})^2$$ as we have observed and that $$\phi((\frac{4}{\ln 10})^2)=g((\frac{4}{\ln 10})^2)-f((\frac{4}{\ln 10})^2)=2(1+\ln (\frac {\ln 10}{4})) >0$$ we have what we want.

• @mathisfun Yes the base of $\log$ is $e$. You shoul derivate $\log a$ and $\sqrt{a} \frac{\log 10}{2}$ to get that $\frac 1 a \leq \frac {1}{\sqrt{a}}\frac {\log 10} {4}$ Jun 14, 2020 at 12:09
• Check the last line it is not correct..in order that the above inequality holds we need $\sqrt{a} > \frac{4}{\ln 10}$ which means $a>{(\frac{4}{\ln 10})}^2$ . Your last line is incorrect. Jun 14, 2020 at 15:48
• @mathisfun I fixed it. Thank you for the remark. Jun 14, 2020 at 16:08

Since, $$t \to \ln t$$ is strictly increasing, it is equivalent to proving $$\ln a < \sqrt{a}\cdot \frac{\ln 10}{2}$$.
Let us define $$\phi (x) =\sqrt{x}\cdot \frac{\ln 10}{2} - \ln x$$ for $$x\geq 2$$.
$$\phi '(x)= \frac {1}{\sqrt{x}}\frac {\ln 10} {4} - \frac{1}{x}.$$ So, $$\phi '(x) >0 \iff \frac {1}{\sqrt{x}}\frac {\ln 10} {4} > \frac{1}{x} \iff \sqrt{x} > \frac{4}{\ln 10}$$
Let, $$d= (\frac{4}{\ln 10})^2$$
Now, for $$x > d,$$ we have $$\sqrt{x} > \frac{4}{\ln 10} \implies \phi '(x) >0 \implies \phi$$ is strictly increasing for $$x>d.$$
Similarly, $$\phi$$ is strictly decreasing for $$2 \leq x i.e. $$\phi$$ attains its minimum at $$x=d.$$
Therefore, for $$x \geq 2,$$ we have $$\phi (x) \geq \phi (d) = 2 -2 \ln (\frac{4}{\ln 10}) >0.$$
This implies, for $$x \geq 2,$$ $$\phi (x) >0 \implies \sqrt{x} \cdot \frac{\ln 10}{2} > \ln x.$$

• I think it is correct. Jun 14, 2020 at 17:00
• Dear downvoter, kindly let me know if there is anything wrong in this proof.. otherwise please don't downvote. Jun 14, 2020 at 20:32